December 23, 2024, 04:29:55 AM
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Topic: Very hard problems abt limiting reagents/empirical formulas/conversions and more  (Read 12964 times)

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Offline SAV_Polite

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I read the rules and I know it looks like a lot of HW problems, but it really isn't! and i have attempted these problems throughout the whole summer and asking online is my last resort. I know you may not be able to answer all these questions ASAP, but even a little help is greatly appreciated! Thank you so much for all your help :) ! I also don't just want the answer. I want to learn so I won't have to ask these questions again. Thank you so much!

1. The reaction between silver ion and solid zinc is represented by the following equation:
2Ag(aq) + Zn(s) >>> Zn^2+(aq) + 2Ag(s)
A 1.50g sample of Zn is combined with 250 mL of 0.110M AgNO3 at 25degreescelcius.
a. identify the limiting reagent. Show calculations to support your answer.
b. On the basis of the limiting reactant that you identified in part a, determine the value of [Zn^2+] after the reaction is complete.

2. The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and investigating its colligative properties.
a. The hydrocarbon burns completely, producing 7.2g of Water and 7.2 liters of CO2 at standard conditions. What is the empirical formula of the hydrocarbon?
b. Calculate the mass in grams of O2 required for the complete combustion of the sample of the hydrocarbon described in a.
c. The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing 100g of CHCl3 and 0.600g of the hydrocarbon is -64.0degreescelcius. The molal freezing point depression constant of CHCl3 is 4.68degreescelcius/molal and its normal freezing point is -63.5degreescelcius. Calculate the molecular mass of the hydrocarbon.
d. What is the molecular formula of the hydrocarbon?

3. Determine the number of protons, electrons and neutrons in each isotope.
a. 151N
b. 147N
c. 23892U
d. 23592U

4. Given the data below, determine the average atomic mass
Isotope % abundance Isotopic Mass
a. Sb-121 57% 120.9038 amu
Sb - 123 42.75% 122.0041 amu
b. Ag-107 51.82% 106.90509 amu
Ag-109 48.18% 108.9047 amu

5. Convert each of the following to moles.
a. 3.00 x 10^24 atoms Au
b. 3.011 x 10^22 molecules H2O

6.Benzene contains only carbon and hydrogen and has molar mass of 78.1g/ mol. Analysis shows the compound to be 7.74% H by mass. Find the empirical and molecular formulas of benzene.

7. Calcium carbonate decomposes upon heating, producing calcium oxide and carbon dioxide gas. How many grams of calcium oxide will be produced after 12.25 g of calcium carbonate is completely decomposed? What volume of carbon dioxide gas is produced from this amount of calcium carbonate, at STP?

8. When ammonia gas, oxygen and methane gas (CH4) are combined. the products are hydrogen cyanide gas and water. Calculate the mass of each product produced when 225g of oxygen gas is reacted with an excess of the other two reactants. If the actual yield of the experiment is 105g of HCN, calculate the percent yield.

Offline Borek

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You have to show your attempts to receive help. This is a forum policy.
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Offline SAV_Polite

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Well, it's too late now. I really needed your help and now I'm going to start chemistry off with a big fat 50% in the records book so thanks a lot for all your help. I'm so sorry I'm not smart enough to answer the questions. This is really frustrating amd stressful for me. I feel for people like me. They really need your help and it really sucks that you won't give them any.

Offline Fridushka

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Let's do some work!
1st of all, for the first problem..
Write the equation , i.e. 2Ag(aq) + Zn(s) >>> Zn^2+(aq) + 2Ag(s)
Use the following information: [quote A 1.50g sample of Zn is combined with 250 mL of 0.110M AgNO3 at 25degreescelcius [/quote] to get your limiting reagent
Hence, try to do some work in order to follow up with the coming questions!

Offline Borek

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i have attempted these problems throughout the whole summer and asking online is my last resort

Well, it's too late now.

Sorry? You had whole summer, you have posted the question on August 30th and 6 hours later it is too late?

Quote
I feel for people like me. They really need your help and it really sucks that you won't give them any.

It really sucks that you don't understand we help in our free time and out of a good will.

I don't feel for people like you. If you didn't know how to solve these questions you could ask a week ago, as you said - you had whole summer. You would be helped in time. You are the only person responsible for the failure.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline DrCMS

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SAV if you don't like the rules then go away and don't bother coming back.
We're here to help people learn how to solve problems not just to do their work for them.  
If we just give you the answers you've learnt nothing and are no better off than you are now.
There are lots of people on the forum who are prepared to help you learn but none of us will just do the work for you.  
If that's too much hard work for you then quit now, drop out of school and die on the streets out of your head on drugs.  
It's your call but don't expect me to cry if you take the easy option.

Offline SAV_Polite

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Thank you Fridushka! Even just leading me into problem helped! :)

SAV if you don't like the rules then go away and don't bother coming back.
We're here to help people learn how to solve problems not just to do their work for them.  
If we just give you the answers you've learnt nothing and are no better off than you are now.
There are lots of people on the forum who are prepared to help you learn but none of us will just do the work for you.  
If that's too much hard work for you then quit now, drop out of school and die on the streets out of your head on drugs.  
It's your call but don't expect me to cry if you take the easy option.

That was just mean with the drugs and everything. That didn't hurt me personally, but if someone were to read that, it could hit them in a soft spot, so you should watch what you say online. And I specifically said i DID NOT just want the answers BECAUSE I wanted to learn so I wouldn't have to ask the questions again. Anyway, school has begun, and I am learning all I need without you guys. I know the second post sounded very mean, but I mean, I was angry at you guys, just sad. Please don't take it the wrong way ...

Offline Fridushka

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You're welcome! Good Luck:)

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